list Prolog，访问列表的特定成员？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/12939425/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Prolog, access specific member of list?
提问by David Carpenter
Can anyone tell me how to access a specific member of a list in prolog? Say for example I need to access the 3rd or 4th element of a list passed into a rule?
谁能告诉我如何访问序言中列表的特定成员？比如说我需要访问传递给规则的列表的第三个或第四个元素？
回答by joel76
nth0(Ind, Lst, Elem)or nth1(Ind, Lst, Elem)with SWI-Prolog, nth0the first element has the index 0.
nth0(Ind, Lst, Elem)或nth1(Ind, Lst, Elem)使用 SWI-Prolog，nth0第一个元素的索引为 0。
For example,
例如，
nth0(3, [a, b, c, d, e], Elem). %Binds d to Elem
nth1(3, [a, b, c, d, e], Elem). %Binds c to Elem
nth0(Ind, [a, b, c, d, e], d). %Binds 3 to Ind
nth0(3, [a, b, c, X, e], d). %Binds d to X
nth0(3, [a, b, c, d, e], c). %Fails.
回答by CapelliC
When the indexes you need to access are so small, you could use pattern matching. Say we need the third element or fourth:
当您需要访问的索引非常小时，您可以使用模式匹配。假设我们需要第三个或第四个元素：
third([_,_,E|_], E).
fourth([_,_,_,E|_], E).
This could be more useful if used 'inline', when the list carries info with positional relevance. For instance
如果使用“内联”，当列表携带具有位置相关性的信息时，这可能更有用。例如
your_rule([_,_,E|Rest], Accum, Result) :-
Sum is Accum + E,
your_rule(Rest, Sum, Result).
...
回答by Nicholas Carey
A prolog list is a classic list. Access is not direct. You have to iterate over to to find what you need.
序言列表是一个经典的列表。访问不是直接的。您必须迭代以找到您需要的内容。
You can get the nth element this way:
您可以通过以下方式获取第 n 个元素：
foo( [X1,X2,X3,X4,...,XN|Xs] ) :- ...
where [code]X[/code]nis the nthelement of the list. Impractical for nlarger than a small value. This is roughly analogous to a C/C++ pointer expression:
其中 [code]X[/code] n是列表的第 n 个元素。对于大于小值的n 是不切实际的。这大致类似于 C/C++ 指针表达式：
LLNode *nthElement = root->next->...->next ;
Otherwise, you have to iterate over the list to find the desired element, using a built-in predicate or a home-brew predicate, something like:
否则，您必须遍历列表以找到所需的元素，使用内置谓词或自制谓词，例如：
foo(Xs) :- nth_element(Xs,9,X) , ...
foo(Xs) :- nth_element(Xs,9,X) , ...
nth_element(Xs,N,X) :- nth_element(Xs,0,N,X) .
nth_element(Xs,N,X) :- nth_element(Xs,0,N,X) 。
nth_element([X|Xs],N,N,X) :- !. nth_element([_|Xs],T,N,X) :- T1 is T+1 , nth_element(Xs,T1,N,X).
nth_element([X|Xs],N,N,X) :- !. nth_element([_|Xs],T,N,X) :- T1 是 T+1 , nth_element(Xs,T1,N,X)。
回答by Anderson Green
Using the funclibrary for SWI-Prolog, it is possible to write list comprehensions in a more concise way:
使用funcSWI-Prolog 库，可以以更简洁的方式编写列表推导式：
:- use_module(library(func)).
nth0((Index, List), Result) :-
nth0(Index,List,Result).
Now, you can access two elements of the list and add them together like this:
现在，您可以访问列表的两个元素并将它们添加在一起，如下所示：
example :-
List = [1,5,12,9],
Y is (nth0 $ (0, List)) + (nth0 $(3,List)), %add the 1st and 4th elements of this list
writeln(Y). %prints 10
